\newproblem{lay:2_1_4}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 2.1.4}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
  Compute $A-5I_3$ and $(5I_3)A$, where $$A=\begin{pmatrix}5 & -1 & 3 \\ -4 & 3 & -6 \\ -3 & 1 & 2\end{pmatrix}$$
}{
   % Solution
	\begin{center}
		$\begin{array}{rcl}
		   A-5I_3&=&\begin{pmatrix}5 & -1 & 3 \\ -4 & 3 & -6 \\ -3 & 1 & 2\end{pmatrix}-5\begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}=
			          \begin{pmatrix}5 & -1 & 3 \\ -4 & 3 & -6 \\ -3 & 1 & 2\end{pmatrix}-\begin{pmatrix}5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{pmatrix}=\begin{pmatrix}0 & -6 & -2 \\ -9 & -2 & -11 \\ -8 & -4 & -3 \end{pmatrix} \\
		   (5I_3)A&=&\left(5\begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\right)\begin{pmatrix}5 & -1 & 3 \\ -4 & 3 & -6 \\ -3 & 1 & 2\end{pmatrix}=
			          \begin{pmatrix}5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{pmatrix}\begin{pmatrix}5 & -1 & 3 \\ -4 & 3 & -6 \\ -3 & 1 & 2\end{pmatrix}=\begin{pmatrix}25 & -5 & 15 \\ -20 & 15 & -30 \\ -15 & 5 & 10 \end{pmatrix} \\
		\end{array}$
	\end{center}
}
\useproblem{lay:2_1_4}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
